Search Results

Search found 41996 results on 1680 pages for 'facebook php sdk'.

Page 224/1680 | < Previous Page | 220 221 222 223 224 225 226 227 228 229 230 231 | Next Page >

Lack of security in many PHP applications?

- by John

Over the past year of freelancing, I inherited two web projects, both of them built in PHP, both of them with sensitive information like credit card info, bank info, etc... In one application, when I typed http://thecompany.com/admin/, and without being asked for a username and password, I saw every user's sensitive information, including credit card numbers, bank account numbers etc... In another application, I was able to bypass the login screen by simply typing http://the2ndcompany.com/customer.php?user_id=777, and again, without any prompts for username and password, i was able to see user 777's credit card info. I cycled through a few more user_ids (any integer) and saw each person's credit card info. Is something wrong here? Or is this the quality of work that the "average" programmer produces? Because if this is what the average programmer produces, does that means I'm an...gasp...elite programmer?? No..that can't be right....something doesn't make sense. So my question is, is it just coincidence that I inherited two applications both of which are dangerously lacking in security? Or are there are a lot of bad PHP programmers out there?

Read the article
GAPI output doesn't match Google Analytics website

- by Yekver

I have to get the main info about my Google Analytics Goals. I'm using GAPI lib, with this code: <?php require_once 'conf.inc'; require_once 'gapi.class.php'; $ga = new gapi(ga_email,ga_password); $dimensions = array('pagePath', 'hostname'); $metrics = array('goalCompletionsAll', 'goalConversionRateAll', 'goalValueAll'); $ga->requestReportData(ga_profile_id, $dimensions, $metrics, '-goalCompletionsAll', '', '2012-09-07', '2012-10-07', 1, 500); $gaResults = $ga->getResults(); foreach($gaResults as $result) { var_dump($result); } cut this code is output: object(gapiReportEntry)[7] private 'metrics' => array (size=3) 'goalCompletionsAll' => int 12031 'goalConversionRateAll' => float 206.93154454764 'goalValueAll' => float 0 private 'dimensions' => array (size=2) 'pagePath' => string '/catalogs.php' (length=13) 'hostname' => string 'www.example.com' (length=13) object(gapiReportEntry)[6] private 'metrics' => array (size=3) 'goalCompletionsAll' => int 9744 'goalConversionRateAll' => float 661.05834464043 'goalValueAll' => float 0 private 'dimensions' => array (size=2) 'pagePath' => string '/price.php' (length=10) 'hostname' => string 'www.example.com' (length=13) What I see on Google Analytics website on Goals URLs page with the same period of date is: Goal Completion Location Goal Completions Goal Value 1. /price.php 9,396 $0.00 2. /saloni.php 3,739 $0.00 As you can see outputs doesn't match. Why? What's wrong?

Read the article
FastCgiModule Error 500 on Windows7 Ultimate + IIS 7.5

- by user63179

I'm running IIS 7.5 on Windows 7 Ultimate. I've installed PHP Version 5.2.14 using Microsoft Web Platform Installer. I've created a virtual directory and a file with which I can browse it and it returns all the PHP information just fine. I'm trying to install MantisBT, and when I copy all the file to my virtual directory and browse index.php I receive the following error detail: Error Summary HTTP Error 500.0 - Internal Server Error The page cannot be displayed because an internal server error has occurred. Detailed Error Information Module FastCgiModule Notification ExecuteRequestHandler Handler PHP_via_FastCGI Error Code 0x00000000 Requested URL http://localhost:80/mantisbt/index.php Physical Path V:\wwwroot\mantisbt\index.php Logon Method Anonymous Logon User Anonymous I've changed these in the php.ini file: fastcgi.impersonate = 1 fastcgi.logging = 0 cgi.fix_pathinfo=1 cgi.force_redirect = 0 The Handler Mappings have this information: Request path: *.php Module: FastCgiModule Executable: C:[Path to PHP installation]\php-cgi.exe Name: PHP_via_FastCGI Thank you for any advice on this!!

Read the article
xcode - defining integer variables

- by Mike

If I have #define myVariable 10 and later I use this as in [self doSomething:myVariable]; Supposing doSomethign method is like - (void) doSomething:(int)myNumber; I receive a warning telling me that a "]" is expected before the ";" why is that and how to solve this? thanks for any help.

Read the article
Loading table sections when using headers

- by Luis Tovar

I cant seem to wrap my head around this. I have googled, and overstacked for hours now looking for examples that i can relate to. What I have is two arrays. The name of my first NSMutableArray is "showDates". I have 3 objects in here. Object 0: "Today, May 20th" Object 1: "Tomorrow, May 21st" Object 2: "Saturday, May 22nd" Then I have my second NSMutableArray named "showTimes" I have about 15 objects in there with strings in each object. ( i hope that makes sense? ) Each object is structured like this: Object 0: showID @"98022" eventID @"833" showTime @"1:30pm" showDate @"Today, May 20th" auditorium @"9" venue @"2991" Object 1: showID @"98222" eventID @"813" showTime @"2:30pm" showDate @"Tomorrow, May 21st" auditorium @"9" venue @"2991" Etc, etc, .... I have the headers working great in my tableView, but I cant seem to figure out how to add the objects in my "showTimes" array under the correct header. Any help would be greatly appreciated.

Read the article
How to cleverly stop "while loop" (php)

- by user3735697

I'm having trouble with creating code that echoes a bunch of stuff that is corresponding to the mysql database row. It needs to keep creating the content until all rows are used and then stop. But for some reason the php file causes the browser to keep loading (it never ends). Any help would be appreciated! Thanks! <?php mysql_connect ("localhost", "root", "") or die ("We couldn't connect!"); mysql_select_db ("dr"); mysql_query ("SELECT * FROM songs"); $result = mysql_query ("SELECT * FROM songs"); while ($row=mysql_fetch_array($result)) { $name = $row ['songname']; $genres = $row ['songgenres']; $mediafire = $row ['mediafirelink']; $dropbox = $row ['dropboxlink']; $source = $row ['audiosource']; echo " <div class='playing'>  <li class='songnameli' id='$source'> <span class='info'>$name</span> <audio> <source src='music/singles/$source.mp3'> <source src='music/singles/$source.ogg'> </audio> </li>  <li class='playlistli'> <img src='icons/addtoplaylist.png' title='Add tot the playlist!' /> </li>  <li class='genresli'> <img src='icons/genres.png' title='Related genres' /> <span class='addedtext genres'>$genres</span> </li>  <li> <span> <img src='icons/share.png' alt='Share this with your friends!' title='Share this!'>  <a href='http://www.facebook.com/sharer.php?u=http://www.declassified-recordings.com' class='addedtext nlink' target='blank_' onclick='popup (this.href, 800, 500); return false'>Facebook </a> <span>/</span>  <a href='http://twitter.com/share? text=Thank%20you%20For%20Sharing!%20It%20means%20the%20world%20to%20us!%40Declassifi3d%20 &url=http://www.declassified-recordings.com' class='twitterlink nlink' target='blank_' onclick='popup (this.href, 800, 500); return false'>Twitter</a> </span> </li>  <li> <img src='icons/download.png' title='Download!' /> <span>  <a href='$mediafire' class='addedtext nlink' target='_blank'>Mediafire</a> <span class='genres'>/</span>  <a href='$mediafire' class='twitterlink nlink' target='_blank'>Dropbox</a> </span> </li> </div>"; } mysql_close (); ?>

Read the article
Initializing Detail View from nib with parameters passed from Root View

- by culov

I'm have a map view with a number of annotations on it... once the callout is clicked, i need to pass several parameters to the DetailViewController, so ive been trying to do this through the constructor. I've debugged a bit and discovered that the arguments are being passed properly and are being received as expected within the constructor, but for some reason whenever I try to change the values of the IBOutlets I've positioned in the nib, it never has an effect. Here's what im passing (btw, im getting a "No initWithNibName : bundle : header' method found" warning at this line): DetailViewController *dvc = [[DetailViewController alloc] initWithNibName:@"DetailViewController" bundle:nil header:headerText]; [self.navigationController pushViewController:dvc animated:YES]; Now heres my constructor: - (id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil header:(UILabel*)headerLabel { if ((self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil])) { self.headerTextView = headerLabel; NSLog(@"header:%@", headerLabel.text); } return self; } Once again, the problem is that headerLabel.text is printed properly in the console, but the line self.headerTextView = headerLabel; doesnt seem to be doing what I want it to do. Thanks!

Read the article
iPhone - Switch Tab from UIWebView

- by user269737

I have a webview and I want to manually switch to another tab during a specific event. The event is arbitray, but I can't figure out hwo to switch teh tab. On my other views, I have the following code: self.tabBarController.selectedViewController = [self.tabBarController.viewControllers objectAtIndex:2]; However, this does not work in my custom UIWebView class. Is there a simple solution to this? Thanks

Read the article
Refactoring or Rewriting Monolithic PHP Spaghetti Codebase

- by nategood

I've inherited a really poorly designed PHP spaghetti code project. It's been gaining a good bit of traffic recently and is starting to have performance issues on top of the poor monolithic code base. Its maxing out performance on a chunky 16GB dedicated machine when it really shouldn't be. I'm planning on doing some performance tweaks right off the bat to help the performance issue, but this still won't really help the horrible code base. The team is small but expecting to grow very soon. I've read Joel's article on the troubles of doing a complete rewrite and see the concerns. But how bad does the code base have to be before you consider a rewrite? There is PHP handling logic interjected into what one would usually consider a "view". Even worse, in some places SQL statements are in these same files! The only real separation of presentation and logic are a few PHP scripts that serve as function libraries. These scripts do most of the ORM stuff... if you can even call it that. Trying to slowly refractor this seems like a nightmare. Open to your thoughts and opinions... however not interested in hearing, "Run away, Run away!".

Read the article
Does ipad have proximity notifications?

- by TheLearner

I assume it doesn't as its only used for when you place the phone near your face correct?

Read the article
SIMPLE PHP MVC Framework [closed]

- by Allen

I need a simple and basic MVC example to get me started. I don't want to use any of the available packaged frameworks. I am in need of a simple example of a simple PHP MVC framework that would allow, at most, the basic creation of a simple multi-page site. I am asking for a simple example because I learn best from simple real world examples. Big popular frameworks (such as code igniter) are to much for me to even try to understand and any other "simple" example I have found are not well explained or seem a little sketchy in general. I should add that most examples of simple MVC frameworks I see use mod_rewrite (for URL routing) or some other Apache-only method. I run PHP on IIS. I need to be able to understand a basic MVC framework, so that I could develop my own that would allow me to easily extend functionality with classes. I am at the point where I understand basic design patterns and MVC pretty well. I understand them in theory, but when it comes down to actually building a real world, simple, well designed MVC framework in PHP, I'm stuck. Edit: I just want to note that I am looking for a simple example that an experienced programmer could whip up in under an hour. I mean simple as in bare bones simple. I don't want to use any huge frameworks, I am trying to roll my own. I need a decent SIMPLE example to get me going.

Read the article
change launch image(default.png)??

- by hemant

i am quite new to iphone development but i have been given a old application and told to change the launch image(default.png)..i tried to google it but couldn't get much help from it?? how can i change the default image in my old application?? any information regarding it will be really helpful??

Read the article
Network reachability on iPhone

- by Infinity

Hello guys! Maybe is there a good way to determine if I have an internet connection or not? (3G or wifi) Thanks!

Read the article
ORDER BY column_name help (via link in HTML table view) (PHP MySQL

- by Derek

My output for my table in HTML has several columns such as userid, name, age, dob. The table heading is simply the title of the column name, I want this to be a link, and when clicked, the selected column is sorted in order, ASC, and then DESC (on next click). I thought this was pretty straight forward but I'm having some difficulty. So far, I have produced this, and no output is taken, apart from the URL works by displaying 'users.php?orderby=userid' <?php if(isset($_GET['orderby'])){ $orderby = $_GET['orderby']; $query_sv = "SELECT * FROM users BY ".mysql_real_escape_string($orderby)." ASC"; } //default query else{ $query_sv = "SELECT * FROM users BY user_id DESC"; } ?> <tr> <th><a href="<?php echo $_SERVER['php_SELF']."?orderby=userid";?>">User ID</a></th> Hoefully if I get this working, I can sort the users by D.O.B. next also using the same principles. Does anyone have any ideas?

Read the article
Execute javascript in PHP

- by Andreas Bonini

I'm generating your typical Web 2.0 HTML page with PHP: it contains a lot of <script> tags and javascript code that will substantially change the DOM after the load event. Is there a way to get the final HTML code directly from PHP, without opening the page with any browser? For example, let's say the HTML for the page is (it's just an example): <html> <head> <script>...the jquery library code...</script> <script>$(document).ready(function() { $("body").append("<p>Hi!</p>");</script> </head> <body> </body> </html> This HTML is saved in the $html PHP variable. Now, I want to pass that variable to some function that will return $result = <html>....<body><p>Hi!</p></body></html>. Is this possible?

Read the article
Getting a UIImage from MySQL using PHP and jSON

- by Daniel

I'm developing a little news reader that retrieves the info from a website by doing a POST request to a URL. The response is a jSON object with the unread-news. E.g. the last news on the App has a timeStamp of "2013-03-01". When the user refreshes the table, it POSTS "domain.com/api/api.php?newer-than=2013-03-01". The api.php script goes to the MySQL database and fetches all the news posted after 2013-03-01 and prints them json_encoded. This is // do something to get the data in an array echo $array_of_fetched_data; for example the response would be [{"title": "new app is coming to the market", "text": "lorem ipsum dolor sit amet...", image: XXX}] the App then gets the response and parses it, obtaining an NSDictionary and adds it to a Core Data db. NSDictionary* obtainedNews = [NSJSONSerialization JSONObjectWithData:responseData options:kNilOptions error:&error]; My question is: How can I add an image to the MySQL database, store it, pass it using jSON trough a POST HTTP Request and then interpret it as an UIImage. It's clear that to store an UIImage in CoreData, they must be transform into/from NSData. How can I pass the NSData back and forth to a MySQL db using php and jSON? How should I upload the image to the db? (Serialized, as a BLOB, etc)

Read the article
Javascript + PHP $_POST array empty

- by Peterim

While trying to send a POST request via xmlhttp.open("POST", "url", true) (javascript) to the server I get an empty $_POST array. Firebug shows that the data is being sent. Here is the data string from Firebug: a=1&q=151a45a150.... But $_POST['q'] returns nothing. The interesting thing is that file_get_contents('php://input') does have my data (the string above), but PHP somehow doesn't recognize it. Tried both $_POST and $_REQUEST, nothing works. Headers being sent: POST /test.php HTTP/1.1 Host: website.com User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv:1.9.2.3) Gecko/20100401 Firefox/3.6.3 Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8 Accept-Language: en-us;q=0.7,en;q=0.3 Accept-Encoding: gzip,deflate Accept-Charset: utf-8;q=0.7,*;q=0.7 Keep-Alive: 115 Connection: keep-alive Referer: http://website.com/ Content-Length: 156 Content-Type: text/plain; charset=UTF-8 Pragma: no-cache Cache-Control: no-cache Thank you for any suggestions.

Read the article
TouchCode XML parsing error

- by itsaboutcode

Hi, I have a xml document which has only one element in the document, which is <error>error string<error> But when i try to parse it, it says this document has no element at all. In other words when i try to access the rootElement it says "null" CXMLDocument *rssParser = [[[CXMLDocument alloc] initWithContentsOfURL:url options:0 error:nil] autorelease]; NSLog(@"Root: %@",[[rssParser rootElement] name]); Please tell me what is wroing with this. Thanks

Read the article
Why is my PHP upload script not working?

- by Turner

Hello all, I am doing some simple work with uploading a file. I am ignoring error checking and exceptions at this point just to get my uploads working. I have this HTML form: <form action='addResult.php' method='post' enctype='multipart/form-data' target='results_iFrame' onsubmit='startUpload();'> Entry: <input type='text' id='entry' /> Stop: <input type='text' id='stop' /> Final: <input type='text' id='final' /> Chart: <input type='file' id='chart' /> <input type='submit' value='Add' /></form> As you can see, it calls 'addResult.php' within the iFrame 'results_iFrame'. The Javascript is just for animation purposes and to tell me when things are finished. addResult.php has this code in it (along with processing the other inputs): $upload_dir = "../img/"; $chart_loc = $upload_dir.basename($_FILES['chart']['name']); move_uploaded_file($_FILES['chart']['tmp_name'], $chart_loc); print_r($_FILES); It uses the 'chart' input from the form and tries to upload it. I have the print_r() function to display some information on $_FILES, but the array is empty, thus making this fail. What could I be doing wrong?

Read the article
PHP Class Question

- by Jerry

Hi all I am trying to build a PHP class to check the username and password in MySql. I am getting "mysql_query(): supplied argument is not a valid MySQL-Link resource in D:\2010Portfolio\football\main.php on line 38" database has errors:" message. When I move my userQuery code out of my Class, it works fine. Only if it is inside my Class. I am not sure if the problem is that I don't build the connection to mysql inside my Class or not. Thanks for any helps!! Here is my code <?php $userName=$_POST['userName']; $userPw=$_POST['password']; class checkUsers { public $userName; public $userPw; function checkUsers($userName='', $userPw=''){ $this->userName=$userName; $this->userPw=$userPw; } function check1(){ if (empty($this->userName) || empty($this->userPw)){ $message="<strong>Please Enter Username and Password</strong>"; }else{ //line 38 is next line $userQuery=mysql_query("SELECT userName, userPw FROM user WHERE userName='$this->userName' and userPw='$this->userPw'", $connection); if (!$userQuery){ die("database has errors: ". mysql_error()); } if(mysql_num_rows($userQuery)==0){ $message="Please enter valid username and password"; } }//end empty check return $message; }//end check1 method }//end Class $checkUser=new checkUsers($userName, $userPw); echo $checkUser->check1(); ?> I appreciate any helps!!!!

Read the article
Finding cause of memory leaks in large PHP stacks

- by Mike B

I have CLI script that runs over several thousand iterations between runs and it appears to have a memory leak. I'm using a tweaked version of Zend Framework with Smarty for view templating and each iteration uses several MB worth of code. The first run immediately uses nearly 8MB of memory (which is fine) but every following run adds about 80kb. My main loop looks like this (very simplified) $users = UsersModel::getUsers(); foreach($users as $user) { $obj = new doSomethingAwesome(); $obj->run($user); $obj = null; unset($obj); } The point is that everything in scope should be unset and the memory freed. My understanding is that PHP runs through its garbage collection process at it's own desire but it does so at the end of functions/methods/scripts. So something must be leaking memory inside doSomethingAwesome() but as I said it is a huge stack of code. Ideally, I would love to find some sort of tool that displayed all my variables no matter the scope at some point during execution. Some sort of symbol-table viewer for php. Does anything like that or any other tools that could help nail down memory leaks in php exist?

Read the article
ipad simulator rotating

- by Mike

I am testing an application on iPad Simulator and I need it to start my app in the position the simulator is but every time I run the app, the simulator rotates the simulator to portrait. Is there a way to stop this behavior? thanks.

Read the article
Filter sub-categories like in layered navigation

- by russjman

I created a new template file catalog/category/list.phtml. This is to display all sub categories of the current category. I have layered navigation which is displaying sub-categories as one of the filters, but I want this new template to work with these filters as well. Right now when i click the subcategory filter, it filters all products on the page, but still displays all categories of the parent category. $_filters is how i am trying to access these filters, but i get nothing. Is there something i am not initializing correctly to have access to these filters from the layered navigation. <?php $_helper = $this->helper('catalog/output'); $_filters = $this->getActiveFilters(); echo $_filters; if (!Mage::registry('current_category')) return ?> <?php $_categories=$this->getCurrentChildCategories() ?> <?php $_count = is_array($_categories)?count($_categories):$_categories->count(); ?> <?php if($_count): ?> <?php foreach ($_categories as $_category): ?> <?php if($_category->getIsActive()): ?> <?php $cur_category=Mage::getModel('catalog/category')->load($_category->getId()); $layer = Mage::getSingleton('catalog/layer'); $layer->setCurrentCategory($cur_category); $_imgHtml = ''; if ($_imgUrl = $this->getCurrentCategory()->getImageUrl()) { $_imgHtml = '<img src="'.$_imgUrl.'" alt="'.$this->htmlEscape($_category->getName()).'" title="'.$this->htmlEscape($_category->getName()).'" class="category-image" />'; $_imgHtml = $_helper->categoryAttribute($_category, $_imgHtml, 'image'); } echo $_category->getImageUrl(); ?> <div class="category-image-box"> <div class="category-description clearfix" > <div class="category-description-textbox" > <h2><span><?php echo $this->htmlEscape($_category->getName()) ?></span></h2> <p><?php echo $this->getCurrentCategory()->getDescription() ?></p> </div> <a href="<?php echo $this->getCategoryUrl($_category) ?>" class="collection-link<?php if ($this->isCategoryActive($_category)): ?> active<?php endif ?>" >See Entire Collection</a> <a href="<?php echo $this->getCategoryUrl($_category) ?>"><?php if($_imgUrl): ?><?php echo $_imgHtml ?><?php else: ?><img src="/store/skin/frontend/default/patio_theme/images/category-photo.jpg" class="category-image" alt="collection" /><?php endif; ?></a> </div> <?php echo '<pre>'.print_r($_category->getData()).'</pre>';?> </div> <?php endif; ?> <?php endforeach ?> <?php endif; ?>

Read the article
Retrieving database column using JSON [migrated]

- by arokia

I have a database consist of 4 columns (id-symbol-name-contractnumber). All 4 columns with their data are being displayed on the user interface using JSON. There is a function which is responisble to add new column to the database e.g (countrycode). The coulmn is added successfully to the database BUT not able to show the new added coulmn in the user interface. Below is my code that is displaying the columns. Can you help me? table.php $(document).ready(function () { // prepare the data var theme = getDemoTheme(); var source = { datatype: "json", datafields: [ { name: 'id' }, { name: 'symbol' }, { name: 'name' }, { name: 'contractnumber' } ], url: 'data.php', filter: function() { // update the grid and send a request to the server. $("#jqxgrid").jqxGrid('updatebounddata', 'filter'); }, cache: false }; var dataAdapter = new $.jqx.dataAdapter(source); // initialize jqxGrid $("#jqxgrid").jqxGrid( { source: dataAdapter, width: 670, theme: theme, showfilterrow: true, filterable: true, columns: [ { text: 'id', datafield: 'id', width: 200 }, { text: 'symbol', datafield: 'symbol', width: 200 }, { text: 'name', datafield: 'name', width: 100 }, { text: 'contractnumber', filtertype: 'list', datafield: 'contractnumber' } ] }); }); data.php <?php #Include the db.php file include('db.php'); $query = "SELECT * FROM pricelist"; $result = mysql_query($query) or die("SQL Error 1: " . mysql_error()); $orders = array(); // get data and store in a json array while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $pricelist[] = array( 'id' => $row['id'], 'symbol' => $row['symbol'], 'name' => $row['name'], 'contractnumber' => $row['contractnumber'] ); } echo json_encode($pricelist); ?>

Read the article
PHP - upload.class image and $_FILES

- by Ockonal

Hello, I'm using class.upload.php to upload pictures onto the server. Here is my form: <form action="<?="http://".$_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'];?>" method="post" enctype="multipart/form-data"> <table style="width: 100%; padding-top: 20px;"> <tr> <td>Image file:</td> <td><input type="file" name="image_file" /></td> </tr> <tr> <td> </td> <td align="right"><input type="submit" name="add_new" value="Add image" /><td></td> </tr> </table> </form> In php code I do: if( array_key_exists('add_new', $_POST) ) { echo 'add new is in array'; echo '<pre>'; print_r($_POST); print_r($_FILES); echo '</pre>'; $handle = new Upload($_FILES['image_file']); ... } Here is an output of print_r: Array ( [image_file] => somefile.png [add_new] => Add image ) Array ( ) As you can see second array is empty ($_FILES), so image doesn't upload. Why? operating system : Linux PHP version : 5.2.12 GD version : 2.0.34 supported image types : png jpg gif bmp open_basedir : /home/httpd/vhosts/kz-gbi.ru/httpdocs:/tmp

Read the article

< Previous Page | 220 221 222 223 224 225 226 227 228 229 230 231 | Next Page >